Question

Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid: (a) 3.00 ml pH- (b) 29.60 mL pH = (C) 38.00 mL pH-

Answer 1

A)when 3.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 3 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.1 M * 3 mL = 0.3 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HBr) = 0.3 mmol

mol(KOH) = 3 mmol

0.3 mmol of both will react

remaining mol of KOH = 2.7 mmol

Total volume = 33.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.7 mmol/33.0 mL

= 8.182*10^-2 M

use:

pOH = -log [OH-]

= -log (8.182*10^-2)

= 1.0872

use:

PH = 14 - pOH

= 14 - 1.0872

= 12.9128

Answer: 12.91

B)when 29.6 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 29.6 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.1 M * 29.6 mL = 2.96 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HBr) = 2.96 mmol

mol(KOH) = 3 mmol

2.96 mmol of both will react

remaining mol of KOH = 4*10^-2 mmol

Total volume = 59.6 mL

[OH-]= mol of base remaining / volume

[OH-] = 4*10^-2 mmol/59.6 mL

= 6.711*10^-4 M

use:

pOH = -log [OH-]

= -log (6.711*10^-4)

= 3.1732

use:

PH = 14 - pOH

= 14 - 3.1732

= 10.8268

Answer: 10.83

C)when 38.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 38 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.1 M * 38 mL = 3.8 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HBr) = 3.8 mmol

mol(KOH) = 3 mmol

3 mmol of both will react

remaining mol of HBr = 0.8 mmol

Total volume = 68.0 mL

[H+]= mol of acid remaining / volume

[H+] = 0.8 mmol/68.0 mL

= 1.176*10^-2 M

use:

pH = -log [H+]

= -log (1.176*10^-2)

= 1.9294

Answer: 1.93